Chapter 12.3, Problem 41PPS

To determine

To find: Value of $\cot \left({270}^0\right)$ .

Answer to Problem 41PPS

  $\cot \left({270}^0\right)=0$

Explanation of Solution

Given information:

  $\cot \left({270}^0\right)$

Concept:

If $\theta$ is the non-quadrantal angle in standard position, its reference angle ${\theta }^{\text{'}}$ is the acute angle formed by the terminal side of $\theta$ and the x-axis.

The ruled for finding the measures of reference angles for $0^0<\theta <{360}^0$ or $0^0<\theta <2\pi$ are shown

For Quadrant − I

  ${\theta }^{\text{'}}=\theta$

For Quadrant − II

  ${\theta }^{\text{'}}={180}^0-\theta$ or ${\theta }^{\text{'}}=\pi -\theta$

For Quadrant − III

  ${\theta }^{\text{'}}=\theta -{180}^0$ or ${\theta }^{\text{'}}=\theta -\pi$

For Quadrant − IV

  ${\theta }^{\text{'}}={360}^0-\theta$ or ${\theta }^{\text{'}}=2\pi -\theta$

Given

  $\cot \left({270}^0\right)$

Here $\theta ={270}^0$

It means

The terminal side of ${270}^0$ lies in Quadrant − IV

Therefore,

  $\begin{array}{l}{\theta }^{\text{'}}={360}^0-\theta \\ \Rightarrow {\theta }^{\text{'}}={360}^0-{270}^0\\ \Rightarrow {\theta }^{\text{'}}={90}^0\end{array}$

Therefore,

  $\cot \left({270}^0\right)=-\cot \left({90}^0\right)$

Because the Cot function is negative in Quadrant − IV

  $\begin{array}{l}\Rightarrow \cot \left({270}^0\right)=-0\\ \Rightarrow \cot \left({270}^0\right)=0\end{array}$

Therefore,

  $\cot \left({270}^0\right)=0$